Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.Scanner;

class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        double dp[] = new double[n + 1];
        dp[0] = 1;
        for(int coin = 0; coin < n; coin++) {
            double p = sc.nextDouble();
            for(int i = coin + 1; i >= 0; i--) {
                dp[i] = ((i == 0) ? 0 : dp[i - 1] * p) + (dp[i] * (1 - p));
            }
        }
        double ans = 0;
        for(int heads = 0; heads <= n; heads++) {
            int tails = n - heads;
            if(heads > tails) {
                ans += dp[heads];
            }
        }
        System.out.println(ans);
    }
}